Let $p: E \to B$ be a Serre fibration, and given a basepoint $b_0 \in B$, let $e_0 \in p^{-1}(b_0)$ be a basepiont in the corresponding fiber.
I'm having trouble proving that the connecting map $\partial_n: \pi_n(B,b_0) \to \pi_{n-1}(p^{-1}(b_0),e_0)$ is a group homomorphism when $n \geq 2$. In the literature, from what I can tell, this is usually done by deriving the long exact sequence of a Serre fibration from the long exact sequence obtained from a suitable pair, so the connecting map in this case is obtained from the analogous map in the sequence of the pair.
I'd like, however, to prove that it's a homomorphism directly, without appealing to the long exact sequence of the pair. Moreover, due to space limitations in the document I'm writing, I'm only dealing with homotopy groups in "spherical" terms, that is, the homotopy group $\pi_n(X,x_0)$ is defined as the set of pointed homotopy classes of pointed maps $[S^n,X]_0$.
Here's an outline of the "spherical" construction of $\partial_n$: given a pointed map $f: S^n \to B$, we consider the composition $f \circ W_{n-1} \circ \pi_{n-1}: S^{n-1} \times I \to B$, where $\pi_{n-1}: S^{n-1} \times I \to \Sigma S^{n-1}$ is the projection to the reduced suspension, and $W_{n-1}: \Sigma S^{n-1} \to S^n$ is the "usual" pointed homeomorphism. The lifting properties of a Serre fibration imply the existence of a homotopy $h^f: S^{n-1} \times I \to E$ satisfying the following properties:
- $h^f(S^{n-1} \times \{0\} \cup \{*_{S^{n-1}}\} \times I) \subseteq \{e_0\}$,
- $p \circ h^f = f \circ W_{n-1} \circ \pi_{n-1}$.
Let's say that a map $S^{n-1} \times I \to E$ satisfying these two properties is a homotopy adapted to $f$.
It follows from these conditions that the "final stage" $h^f_1: S^{n-1} \to E$ of $h^f$ actually takes values in the fiber $p^{-1}(b_0)$, and we define $\partial_n([f]) = [h^f_1] \in \pi_{n-1}(p^{-1}(b_0),e_0)$. I have managed to prove that this construction is well-defined, i.e., that it is independent of the chosen homotopy $h^f$ adapted to $f$ and also independent of the pointed homotopy class of $f$. I have also already proved that $\partial_n$ fits in a long exact sequence of pointed sets.
I got really stuck, however, when trying to prove that $\partial_n$ is a group homomorphism. I'm not even sure which strategy I should use to prove this. Here's an idea that might be worth something.
Given elements $[f], [g] \in \pi_n(B,b_0)$, the product $[f] \cdot [g]$ is represented by the pointed map $\langle f,g \rangle \circ \mu_{S^n}: S^n \to B$, where $\mu_{S^n}: S^n \to S^n \vee S^n$ is the H-comultiplication on the sphere induced from the usual H-comultiplication $\mu_{n-1}: \Sigma S^{n-1} \to \Sigma S^{n-1} \vee \Sigma S^{n-1}$ on the reduced suspension via the homeomorphism $W_{n-1}$.
If $h: S^{n-1} \times I \to E$ is adapted to $\langle f,g \rangle \circ \mu_{S^n}$, then unpacking the definition of $\mu_{S^n}$ one can show that the equality $p \circ h = \langle f \circ W_{n-1}, g \circ W_{n-1} \rangle \circ \mu_{n-1} \circ \pi_{n-1}$ holds.
It follows that $h^f: S^{n-1} \times I \to E$ defined as $h^f(x,t) = h(x,t/2)$ (the "lower part" of $h$) is adapted to $f$, therefore we can write $\partial_n([f]) = [h^f_1] = [h_{1/2}]$.
I don't know how to obtain a homotopy adapted to $g$ directly from $h$. What we can do is consider $h': S^{n-1} \times I \to E$ adapted to $\langle g,f \rangle \circ \mu_{S^n}$ instead, and set $h^g(x,t) = h'(x,t/2)$, which allows us to write $\partial_n([g]) = [h^g_1] = [h'_{1/2}]$.
Since $\pi_n(B,b_0)$ is abelian (because $n \geq 2$), we have the equalities $[h_1] = \partial_n([f] \cdot [g]) = \partial_n([g] \cdot [f]) = [h_1']$ in $\pi_{n-1}(p^{-1}(b_0),e_0)$, so there exists a pointed homotopy $\Theta: S^{n-1} \times I \to p^{-1}(b_0)$ joining $h_1$ to $h_1'$.
So now we have to prove that $h_1 \simeq_* \langle h_{1/2}, h_{1/2}' \rangle \circ \mu_{S^{n-1}}$ or $h_1' \simeq_* \langle h_{1/2}, h_{1/2}' \rangle \circ \mu_{S^{n-1}}$, taking into account that these homotopies must hold inside the fiber $p^{-1}(b_0)$.
I don't know how to obtain such a homotopy. My guess is that it can be obtained from a suitable use of the lifting properties of the fibration. For example, since $S^{n-1} \times I$ can be obtained from the subspace $(S^{n-1} \times \{0\}) \cup (S^{n-1} \times \{1\}) \cup (\{*_{S^{n-1}}\} \times I)$ by attaching cells, a homotopy $\varphi: (S^{n-1} \times I) \times I \to B$ can be lifted to $\widetilde{\varphi}: (S^{n-1} \times I) \times I \to E$ as long as we can lift it over the subspace $$ A = ((S^{n-1} \times I) \times 0) \cup ((S^{n-1} \times 0) \times I) \cup ((S^{n-1} \times 1) \times I) \cup (( * \times I) \times I), $$ but I wasn't able to apply this lifting property. I imagine the homotopy $\Theta$ between $h_1$ and $h_1'$ will be used here somehow.
So, how should I construct the necessary homotopy? Or is there another approach to the problem of showing that $\partial_n$ is a homomorphism?